Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
8. Systems of Linear and Quadratic Equations
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Exercise 37 Page 601

Determine the points of intersection using the parabola's equation. Use these points to write the equation of the line in slope-intercept form.

B

Practice makes perfect

We are asked to write an equation of the line bounding the tabletop. To do so, let's find the coordinates of the points of intersection of the line and the parabola and then use them to write the equation of the line.

Points of Intersection

We know that the parabola and the line intersect when x= - 1 and when x= 3. The points of intersection are the points common to both the parabola and the line. Therefore, to find the y-coordinates of these two points we will use the function that models the parabola. y=2x^2-3x+2 First, let's determine the y-value for x= - 1!

y=2x^2-3x+2
y=2( - 1)^2-3( - 1)+2
â–¼
Simplify
y=2(1)-3(- 1)+2
y=2(1)+3+2
y=2+3+2
y=7
The first point of intersection is (- 1,7). Let's find the other one!

y=2x^2-3x+2
y=2( 3)^2-3( 3)+2
â–¼
Simplify
y=2(9)-3(3)+2
y=18-9+2
y=11

The second point of intersection is (3,11).

Equation of the Line

Recall that exactly one line passes through two points. Therefore, given two points of intersection we are able to write an equation of the line. We will write the equation of the line in slope-intercept form. y=mx+b Here, m is the slope and b is the y-intercept. Let's use the Slope Formula to find the slope m of the line bounding the table. m=y_2-y_1/x_2-x_1 We will use the points of intersection, ( - 1,7) and ( 3,11), as the points (x_1,y_1) and (x_2,y_2).

m=y_2-y_1/x_2-x_1
m=11- 7/3-( - 1)
â–¼
Subtract terms
m=11-7/3+1
m=4/4
m=1

The slope is equal to 1. Let's write a partial equation of the line. y=1x+b ⇔ y=x+b Now we will use one of the points to write an equation that can be solved for b. Let's use ( - 1, 7)!

y=x+b
7= - 1+b
8=b
b=8

Finally, we can finish writing the equation of the line. y=x+8 Our answer corresponds to option B.